Problem: $f(x, y) = (x^{1.5}y, 2x^2y^4)$ What is the curl of $f$ at $(4, 1)$ ?
Solution: The formula for curl in two dimensions is $\text{curl}(f) = \dfrac{\partial Q}{\partial x} - \dfrac{\partial P}{\partial y}$, where $P$ is the $x$ -component of $f$ and $Q$ is the $y$ -component. Let's differentiate! $\begin{aligned} \dfrac{\partial Q}{\partial x} &= \dfrac{\partial}{\partial x} \left[ 2x^2y^4 \right] \\ \\ &= 8xy^4 \\ \\ \dfrac{\partial P}{\partial y} &= \dfrac{\partial}{\partial y} \left[ x^{1.5}y \right] \\ \\ &= x^{1.5} \end{aligned}$ Therefore: $\text{curl}(f) = 4xy^4 - x^{1.5}$ The curl of $f$ at $(4, 1)$ is $8$.